Common fixed point results in metric spaces endowed with a graph

In this paper we establish some common fixed point results for a pair of mappings in a metric space having the additional structure of a directed graph. The mappings are assumed to satisfy certain almost G-contractions without and with rational terms. Each of our results is illustrated with example. The approach here is a blending of analytic and graph theoretic methodologies.


Introduction
The consideration of partial ordering in metric fixed point theory appeared first in the work of Turinici [25] who worked out some results in metrizable uniform spaces.Subsequent work was carried out in this direction by Ran and Reurings [23] where they combined the Banach contraction principle and the Knaster-Tarski fixed point theorem.Ran and Reurings also combined their result with the Schauder fixed point theorem and applied it to obtain some existence and uniqueness results to nonlinear matrix equations.Theorem 2.2.Let (X, d) be a complete metric space endowed with a directed graph G and S, T : X −→ X be two mappings.Suppose that (i) (Sx, TSx) and (Tx, STx) ∈ E(G) for all x ∈ X and (ii) there exist α ∈ (0, 1) and L ≥ 0 such that for all x, y ∈ X with (x, y) ∈ E(G) or (x, y) ∈ E(G −1 ), d(Sx, Ty) ≤ α max d(x, y), d(x, Sx), d(y, Ty), d(x, Ty) + d(y, Sx) 2 + L min d(x, Sx), d(y, Ty), d(x, Ty), d(y, Sx) .Also suppose that (a) S or T is continuous, or (b) the triple (X, d, G) is regular.Then S and T have a common fixed point in X.
Proof.First we prove that any fixed point of S is also a fixed point of T and conversely, any fixed point of T is also a fixed point of S.
which is a contradiction.Hence p = T p, that is, p is a fixed point of T. Using a similar argument, we have that any fixed point of T is also a fixed point of S.
Let S be continuous.By ( 1) and ( 9), we have Since S is continuous, using (10) we have Therefore, z is a fixed point of S. By what we have already proved, z is also a fixed point of T. Hence z is a common fixed point of S and T.
Similarly, if T is continuous, then z is a common fixed point of S and T.
Suppose that the condition (b) holds.
Since {x n } is a sequence in X such that x n −→ z and (x n , x n+1 ) ∈ E(G) for all n ≥ 1, using the regular property of (X, d, G), we have Therefore, Since (x 2n , z) ∈ E(G) for all n ≥ 1, applying the condition (ii), we have Taking the limit as n −→ ∞ in the above inequality and using (9), we have Since α < 1, the above inequality leads to a contradiction unless d(z, Tz) = 0, that is, z = Tz, that is, z is a fixed point of T. By what we have already proved, z is also a fixed point of S. Hence z is a common fixed point of S and T.
In the following theorem mappings are assumed to satisfy a almost G-contraction of rational type.Theorem 2.3.Let (X, d) be a complete metric space endowed with a directed graph G and S, T : X −→ X be two mappings.Suppose that (i) (Sx, TSx) and (Tx, STx) ∈ E(G) for all x ∈ X and (ii) there exist α ∈ (0, 1) and L ≥ 0 such that for all x, y ∈ X with (x, y) ∈ E(G) or which implies that d(p, T p) = 0, that is, p = T p, that is, p is a fixed point of T. Using a similar argument, we have that any fixed point of T is also a fixed point of S.
Arguing similarly as in the proof of the Theorem 2.2 we prove that z is a common fixed point of S and T.
Suppose that the condition (b) holds.
Applying similar logic as described in Theorem 2.2, we have (x 2n , z) ∈ E(G) for all n ≥ 1.Then applying the condition (ii), we have Taking the limit as n −→ ∞ in the above inequality and using (9), we have d(z, Tz) ≤ 0, which implies that d(z, Tz) = 0, that is, z = Tz, that is, z is a fixed point of T. By what we have already proved, z is also a fixed point of S. Hence z is a common fixed point of S and T.

(B)
Let L ≥ 256 any real real number and 1 2 ≤ α < 1.Then all the conditions of Theorem 2.3 are satisfied and here B ∪ { 1 16 } is the set of common fixed points of S and T.
Now suppose that p is a fixed point of S and p = T p. Since (p, p) ∈ E(G), applying the condition (ii), we have d(p, T p) = d(Sp, T p) ≤ α max d(p, p), d(p, Sp), d(p, T p), d(p, T p) + d(p, Sp) 2 + L min d(p, Sp), d(p, T p), d(p, T p), d(p, Sp) ≤ α max d(p, p), d(p, p), d(p, T p), d(p, T p) + d(p, p) 2 + L min d(p, p), d(p, T p), d(p, T p), d(p, p) (x, y) ∈ E(G −1 ), d(Sx, Ty) ≤ α max d(x, y), d(x, y) [1 + d(x, Ty) d(y, Sx)] 1 + d(x, y) + L d(x, Ty) d(y, Sx).Also suppose that (a) S or T is continuous, or (b) the triple (X, d, G) is regular.Then S and T have a common fixed point in X. Proof.Like in the proof of Theorem 2.2 we first prove that any fixed point of S is also a fixed point of T and conversely.Now suppose that p is a fixed point of S and p = T p. Since (p, p) ∈ E(G), applying the condition (ii), we have d(p, T p) = d(Sp, T p) ≤ α max d(p, p), d(p, p) [1 + d(p, T p) d(p, Sp)] 1 + d(p, p) + L d(p, T p) d(p, Sp)

Example 2 . 4 .Let L ≥ 0 any real number and 1 8 ≤ α < 1 .
Let X = A B, where A = [0, 1] and B = {n : n ∈ N and n ≥ 2}.Let G be a directed graph withV(G) = X and E(G) = {(x, y) : x, y ∈ A and x ≤ y} {(n, n) : n ∈ B}.Let d be the usual metric on X.Let S, T : X −→ X be defined respectively as follows: Then all the conditions of Theorem 2.2 are satisfied and here B ∪ {1  16 } is the set of common fixed points of S and T.